15th March 2015 , 7 min read
On Differential Forms
Original post is here eklausmeier.goip.de/blog/2015/03-15-on-differential-forms-2 .
Abstract. This article will give a very simple definition of $k$-forms or differential forms. It just requires basic knowledge about matrices and determinants. Furthermore a very simple proof will be given for the proposition that the double outer differentiation of $k$-forms vanishes.
MSC 2010: 58A10
1. Basic definitions
We denote the submatrix of $A=(a_{ij})\in R^{m\times n}$ consisting of the rows $i_1,\ldots,i_k$ and the columns $j_1,\ldots,j_k$ with
$$
[A]{\textstyle\!{\scriptstyle j_1\atop\scriptstyle i_1}
\!{\scriptstyle\ldots\atop\scriptstyle\ldots}
\!{\scriptstyle j_k\atop\scriptstyle i_k}} := \begin{pmatrix}
a_{i_1j_1} & \ldots & a_{i_1j_k}\\
\vdots & \ddots & \vdots\\
a_{i_kj_1} & \ldots & a_{i_kj_k}\\
\end{pmatrix}
$$
and its determinant with
$$
A{\textstyle\!{\scriptstyle j_1\atop\scriptstyle i_1}
\!{\scriptstyle\ldots\atop\scriptstyle\ldots}
\!{\scriptstyle j_k\atop\scriptstyle i_k}}
:= \det [A]{\textstyle\!{\scriptstyle j_1\atop\scriptstyle i_1}
\!{\scriptstyle\ldots\atop\scriptstyle\ldots}
\!{\scriptstyle j_k\atop\scriptstyle i_k}}.
$$
For example
$$
A = \begin{pmatrix}a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ \end{pmatrix},
\qquad A_{1,2}^{1,3} = a_{11}a_{23} - a_{21}a_{13}.
$$
Suppose
$$
H \in R^{n\times(n+1)}
$$
and let
$$
f,g\colon U\subseteq R^n\to R, \qquad U \text{ open},
$$
be two functions which are two-times continuously differentiable. Then we call for a fixed $k$ the expression
$$
f\,H_\alpha^{1\ldots k},
\qquad \alpha=\left(i_1,\ldots,i_k\right)
\in\left\{1,\ldots,n\right\}^k,
$$
a basic $k$-form or basic differential form of order $k$. It's a real function of $n+k^2$ variables. For $k>n$ the expression is defined to be zero. If $f$ also depends on $\alpha$ then
$$
\sum_{1\le i_1\lt \cdots\lt i_k\le n} f_{i_1\ldots i_k}
H{\textstyle\!{\scriptstyle1\atop\scriptstyle i_1}
\!{\scriptstyle\ldots\atop\scriptstyle\ldots}
\!{\scriptstyle k\atop\scriptstyle i_k}}
$$
is called a $k$-form . It's a real function of $n+kn$ variables which is $k$-linear in the $k$ column-vectors of $H$.
For example for $f\colon R\to R$ and $H\in R^{1\times1}$ we have $f(x),H$. This is a linear function in $H$ and a possibly non-linear function in $x$.
2. Differentiation of $k$-forms
For the differential form
$$
\omega = f H^{1\ldots k}_\alpha,
\qquad \alpha=\left(i_1,\ldots,i_k\right),
$$
we define
$$
d\omega := \sum_{\nu=1}^n {\partial f\over\partial x_\nu}
H^{1\ldots k+1}_{\nu,\alpha}
$$
as the outer differentiation of $\omega$. This is a $(k+1)$-form. It's a function of $n+(k+1)n$ variables.
The $0$-form
$$
\omega = f, \qquad \left|\alpha\right|=k=0
$$
yields
$$
dw = \sum_{\nu=1}^n {\partial f\over\partial x_\nu} H^1_\nu, \tag 1
$$
which corresponds to $\nabla f = \mathop{\rm grad}f$.
In the special case $k=\left|\alpha\right|=1$ we get for
$$
\omega = \sum_{i=1}^n f_i H^1_i
$$
the result
$$
d\omega = \sum_{i=1}^n \sum_{j=1}^n {\partial f_i\over\partial x_j}
H^{1,2}_{j,i}
= \sum_{i\lt j} \left({\partial f_i\over\partial x_j}
- {\partial f_j\over\partial x_i}\right) H^{1,2}_{j,i}. \tag 2
$$
This corresponds to $\mathop{\rm rot} f$.
Let hat ($\hat{}$) mean exclusion from the index list. The case $k=n-1$ for
$$
\omega = \sum_{i=1}^n (-1)^{i-1}
f_i\,H^{\: 1\ldots n-1\: }_{1\ldots\hat\imath\ldots n}
$$
delivers
$$
dw = \sum_{i=1}^n \sum_{\nu=1}^n (-1)^{i-1}
{\partial f_i\over\partial x_\nu}
H^{\: \: 1\ldots n}_{\nu,1\ldots\hat\imath\ldots n}
= \sum_{i=1}^n {\partial f_i\over\partial x_\nu}
H^{1\ldots n}_{1\ldots n}
= \left(\sum_{i=1}^n{\partial f_i\over\partial x_i}\right) \det H.
$$
This corresponds to $\mathop{\rm div}f$.
Theorem. For $\omega = f H_\alpha^{1\ldots k}$ we have
$$
dd\omega = 0.
$$
Proof: With
$$
d\omega = \sum_{\nu=1}^n {\partial f\over\partial x_\nu}
H_{\nu,\alpha}^{1\ldots k+1}
$$
we get
$$
dd\omega = \sum_{\nu=1}^n \sum_{\mu=1}^n
{\partial^2 f\over\partial x_\nu\partial x_\mu}
H_{\mu,\nu,\alpha}^{1\ldots k+2}
$$
and this is zero, because
$$
H_{\mu,\mu,\alpha}^{1\ldots k+2} = 0, \qquad
H_{\mu,\nu,\alpha}^{1\ldots k+2} = -H_{\nu,\mu,\alpha}^{1\ldots k+2},
$$
and
$$
{\partial^2f\over\partial x_\nu\partial x_\mu} =
{\partial^2f\over\partial x_\mu\partial x_\nu}.
$$
Application of this theorem to an $0$-form with an $f\colon U\subseteq R^n\to R$ and a $1$-form with an $a\colon U\to R^n$ reading (1) and then (2) yields
$$
\mathop{\rm rot}\mathop{\rm grad} f = 0,
\qquad \mathop{\rm div}\mathop{\rm rot} a = 0.
$$
The second equation is only true for $n=3$ because
$$
{n\choose 2} = n \quad (n\in N)
\qquad\Leftrightarrow\qquad n = 3.
$$
Definition. Suppose
$$
\phi\colon D\to E\subset R^n, \qquad D\subset\!\subset R^k,
$$
is differentiable, its derivative denoted by $\phi'$, and
$$
f\colon E\to R.
$$
For the differential form $\omega = f H^{1\ldots k}_\alpha$ we define the back-transportation as
$$
\phi^*\omega := (f\circ\phi) \, (\phi')^{1\ldots k}_{\alpha}
$$
and the integral over $k$-forms as
$$
\int_\phi \omega := \int_D \phi^*\omega.
$$
For example the case $k=1$,
$$
\omega = \sum_{i=1}^n f_i H^1_i
$$
gives
$$
\phi^*\omega = \sum_{i=1}^n (f_i\circ\phi) \, (\phi')_i^1 .
$$
3. The outer product of differential forms
Suppose
$$
H\in R^{n\times(n+1)}, \qquad k+m\leq n.
$$
For the two differential forms
$$
\omega = \sum_{1\le i_1\lt \cdots\lt i_k\le n} f_{i_1\ldots i_k}
H{\textstyle\!{\scriptstyle1\atop\scriptstyle i_1}
\!{\scriptstyle\ldots\atop\scriptstyle\ldots}
\!{\scriptstyle k\atop\scriptstyle i_k}}
$$
and
$$
\lambda = \sum_{1\le j_1\lt \cdots\lt j_m\le n} g_{j_1\ldots j_m}
H{\textstyle\!{\scriptstyle k+1\atop\scriptstyle j_1}
\!{\scriptstyle\ldots\atop\scriptstyle\ldots}
\!{\scriptstyle k+m\atop\scriptstyle j_m}}
$$
the outer product is defined as
$$
w\land\lambda :=
\sum
_{\scriptstyle1\le i_1\lt \cdots\lt i_k\le n\atop
\scriptstyle1\le j_1\lt \cdots\lt j_m\le n}
f_{i_1\ldots i_k} g_{j_1\ldots j_m}
H{\textstyle\!{\scriptstyle1\atop\scriptstyle i_1}
\!{\scriptstyle\ldots\atop\scriptstyle\ldots}
\!{\scriptstyle k\atop\scriptstyle i_k}
\!{\scriptstyle k+1\atop\scriptstyle j_1}
\!{\scriptstyle\ldots\atop\scriptstyle\ldots}
\!{\scriptstyle k+m\atop\scriptstyle j_m}} .
$$
This is a differential form of order $k+m$. It's a function in $n+(k+m)n$ variables.
Theorem.
$$
d(\omega\land\lambda) =
d\omega\land\lambda + (-1)^k\omega\land d\lambda
$$
Proof: With
$$
\omega = \sum_\alpha f_\alpha H_\alpha^{1\ldots k}, \qquad
\lambda = \sum_\beta g_\beta H_\beta^{1\ldots m}
$$
then
$$
\eqalign{
d(\omega\land\lambda) &= \sum_{\alpha,\beta} \sum_{\nu=1}^n \left(
{\partial f_\alpha\over\partial x_\nu} g_\beta
+ f_\beta {\partial g_\beta\over\partial x_\nu}
\right) H_{\nu,\alpha,\beta}^{1\ldots k+m+1} \cr
&= \sum_{\alpha,\beta} \sum_{\nu=1}^n
{\partial f_\alpha\over\partial x_\nu} g_\beta
H_{\nu,\alpha,\beta}^{1\ldots k+m+1}
+ \sum_{\alpha,\beta} \sum_{\nu=1}^n
f_\alpha {\partial g_\beta\over\partial x_\nu}
H_{\nu,\alpha,\beta}^{1\ldots k+m+1} \cr
&= d\omega\land\lambda + (-1)^k\omega\land d\lambda,\cr
}
$$
due to
$$
H_{\nu,\alpha,\beta}^{1\ldots k+m+1}
= (-1)^k H_{\nu,\beta,\alpha}^{1\ldots k+m+1}
$$
and
$$
d\lambda = \sum_\beta \sum_{\nu=1}^n
{\partial g_\beta\over\partial x_\nu}
H_{\nu,\beta}^{1\ldots m+1} .
$$
An alternative definition for the differentiation of $k$-forms could be given.
Theorem. Suppose
$$
\omega = f H_\alpha^{1\ldots k}, \qquad0\le\left|\alpha\right|\le k,
$$
and
$$
H = \left(h_1,\ldots,h_n,h_{n+1}\right) \in R^{n\times(n+1)}
$$
with $\alpha=\left(i_1,\ldots,i_k\right)$ we have
$$
d\omega = \det\left(
\mathop{\rm col}\left(
\nabla f,
[{\rm Id}_n]_\alpha^{1\ldots n}
\right) [H]_{1\ldots n}^{1\ldots k+1}
\right) = \sum_{\nu=1}^n {\partial f\over\partial x_\nu}
H_{\nu,\alpha}^{1\ldots k+1},
$$
where $\rm col$ just stacks matrices one above another and ${\rm Id}_n$ is the identity matrix in $R^n$.
Proof:
$$
d\omega =
\left|\begin{matrix}
\left\langle\nabla f,h_1\right\rangle & \ldots & \left\langle\nabla f,h_k\right\rangle & \left\langle\nabla f,h_{k+1}\right\rangle \\
\left\langle e_{i_1},h_1\right\rangle & \ldots & \left\langle e_{i_1},h_k\right\rangle & \left\langle e_{i_1},h_{k+1}\right\rangle \\
\vdots & \ddots & \vdots & \vdots\cr
\left\langle e_{i_k},h_1\right\rangle & \ldots & \left\langle e_{i_k},h_k\right\rangle & \left\langle e_{i_k},h_{k+1}\right\rangle \cr
\end{matrix}\right|
$$
$$
\qquad = \sum_{\nu=1}^n {\partial f\over\partial x_\nu}
\left|\begin{matrix}
h_{1,\nu} & h_{1,i_1} & \ldots & h_{1,i_k}\\
\vdots & \vdots & \ddots & \vdots\\
h_{k,\nu} & h_{k,i_1} & \ldots & h_{k,i_k}\\
h_{k+1,\nu}& h_{k+1,i_1} & \ldots & h_{k+1,i_k}\\
\end{matrix}\right|
$$
since
$$
\left\langle\nabla f,h_1\right\rangle = \sum_{\nu=1}^n {\partial f\over\partial x_\nu} h_{1,\nu},
$$
$$
\vdots\qquad\qquad\vdots
$$
$$
\left\langle\nabla f,h_{k+1}\right\rangle = \sum_{\nu=1}^n {\partial f\over\partial x_\nu} h_{k+1,\nu}.
$$
REFERENCES.
Walter Rudin, Principles of Mathematical Analysis, Second Edition, McGraw-Hill, New York, 1964
Otto Forster, Analysis 3: Integralrechnung im $R^n$ mit Anwendungen, Third Edition, Friedrich Vieweg & Sohn, Braunschweig/Wiesbaden, 1984